\(D=0.7\cdot2\dfrac{2}{3}\cdot20\cdot0.375\cdot\dfrac{5}{28}\)

\(=\dfrac{7}{10}\cdot2\dfrac{2}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\left(\dfrac{8}{3}\cdot\dfrac{3}{8}\right).\left(\dfrac{7}{10}\cdot20\right)\cdot\dfrac{5}{28}\)

\(=\) \(1\) \(\cdot\) \(14\) \(\cdot\) \(\dfrac{5}{28}\)

\(=14\cdot\dfrac{5}{28}\)

\(=\dfrac{5}{2}\).

\(A=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{-5}{7}\)

\(=\dfrac{-5}{7}+\dfrac{-5}{7}+1=\dfrac{-3}{7}\)

\(B=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(=\dfrac{7}{10}.\dfrac{5}{28}.20=\dfrac{5}{2}.\)

Ta có :

A = \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

= \(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}+1\)

= \(\left(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}\right)+1\)

= \(\dfrac{5}{7}.\left(\dfrac{-2}{11}-\dfrac{9}{11}+1\right)+1\)

= \(\dfrac{5}{7}.0+1\)

= 1

B = \(0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

= \(\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

= \(\left(\dfrac{7}{10}.20\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\dfrac{5}{28}\)

= 14.1.\(\dfrac{5}{28}\)

= \(\dfrac{5}{2}\)

Vậy A = 1

B = \(\dfrac{5}{2}\)

7/10 * 8/3 * 20 * 3/8 * 5/28

= (7/10 * 20) * (8/3 * 3/8) * 5/28

= 14 * 1 * 5/28

= 14 * 5/28

= 5/2

0,7. 2\(\dfrac{2}{3}\) . 20. 0,375. \(\dfrac{5}{28}\) 

= \(\dfrac{14}{15}\) . 20. \(\dfrac{15}{224}\) 

= \(\dfrac{56}{3}\) . \(\dfrac{15}{224}\) = \(\dfrac{5}{4}\)

= 7/10 x 8/3 x 20 x 375/1000 x 5/28
= 7/10 x 8/3 x 20 x  3/8 x 5/28 
= ( 7/10 x20)x(8/3x3/8)x5/28
=14 x 1 x 5/28
= 14 x 5/28
= 5/2

\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)

\(A=6-2\dfrac{4}{7}\)

\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)

\(A=3\dfrac{3}{7}\)

\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)

\(B=2+3\dfrac{7}{11}\)

\(B=5\dfrac{7}{11}\)

\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{160}{11}\)

\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)

\(D=\dfrac{7}{28}=\dfrac{5}{2}\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)

\(\Rightarrow E=0\)

3,5

đổi số thập phân thành phân số.

Bài 1:

a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{2}{4}\)

\(=\dfrac{3}{4}\)

b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)

\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)

\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

\(=\dfrac{9}{5}\)

c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)

\(=\dfrac{1}{6}-\dfrac{7}{2}\)

\(=\dfrac{1}{6}-\dfrac{21}{6}\)

\(=\dfrac{-10}{3}\)

e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\dfrac{2}{3}\)

f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{3}{2}\)

\(=\dfrac{2}{2}=1\)

g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)

\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{2}{4}-\dfrac{3}{4}\)

\(=\dfrac{-1}{4}\)

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{9}{28}\)

\(=\dfrac{196}{140}-\dfrac{45}{140}\)

\(=\dfrac{151}{140}\)

i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)

\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)

\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)

\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)

k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)

\(=-\dfrac{2}{3}\)

\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)

\(A=\dfrac{1}{8}.1.20\)

\(A=\dfrac{20}{8}=\dfrac{5}{2}\)

\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)

\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)

\(B=\left(16+1\right)+4,03\)

\(B=17+4,03\)

\(B=21,03\)

\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)

\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)

\(C=390.\dfrac{15}{78}\)

\(C=75\)

a) \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

\(=\dfrac{1135}{23}-\left(\left(5+14\right)+\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\right)\)

\(=\dfrac{1135}{23}-\left(19+\dfrac{417}{736}\right)\)

\(=\dfrac{1135}{23}-19\dfrac{417}{736}\)

\(=\dfrac{1135}{23}-\dfrac{14401}{736}\)

\(=\dfrac{953}{32}\)

b) \(-\dfrac{3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=-\dfrac{1}{7}\cdot\dfrac{5}{3}-\dfrac{4}{3}\cdot\dfrac{1}{7}+\dfrac{17}{7}\)

\(=-\dfrac{5}{21}-\dfrac{4}{21}+\dfrac{17}{7}\)

\(=2\)

c) \(0,7\cdot2\dfrac{2}{3}\cdot20\cdot0,375\cdot\dfrac{5}{28}\)

\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=2\cdot\dfrac{5}{4}\)

\(=\dfrac{5}{2}\)

d) \(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(=\left(9\cdot\dfrac{3}{8}+\dfrac{303030}{69264}\right)+\dfrac{403}{100}\)

\(=\left(\dfrac{27}{8}+\dfrac{35}{8}\right)+\dfrac{403}{100}\)

\(=\dfrac{31}{4}+\dfrac{403}{100}\)

\(=\dfrac{589}{50}\)

P/s: Đánh dấu phẩy, dấu chấm (dấu nhân) cần rõ ràng (vì dấu chấm người ta sẽ hiểu là dấu nhân thay vì hiểu là dấu phẩy)

a) \(49\dfrac{8}{23}\)- \(\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

= \(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)+5\dfrac{7}{32}\)

=35+\(5\dfrac{7}{32}\)

=\(\dfrac{1287}{32}\)

b)\(-\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).\dfrac{9}{9}\right]+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).1\right]+2\dfrac{3}{7}\)

=\(\left(\dfrac{-3}{7}\right)+2\dfrac{3}{7}\)

=2

c) 0,7.\(2\dfrac{2}{3}\).20.0.375.\(\dfrac{5}{28}\)

=0 (Vì có một thừ số là 0 nên nguyên cả tích là 0)

d)\(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

=17+4,03

=21,03

Ta có : \(D=\frac{1598}{66}\)

\(E=\frac{5656}{55}\)

E=(-0,17+3+5/9-2-36/97).(1/12-1/12)=(-0,17+3+5/9-2-36/97).0=0